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Integral of higher number ∫ DX / [x & # 178; (1 + X & # 178;)] please write in detail
The denominator is split into two terms to subtract: ∫ DX / X & # 178; (1 + X & # 178;) = ∫ [(1 / X & # 178;) - 1 / (1 + X & # 178;)] DX = - (1 / x) + arctanx + C
Integration of higher numbers ∫ 1 / [x √ (4-x & # 178;)] DX
Finding indefinite integral ∫ 1 / [x √ (4-x & # 178;)] DX
The original formula = (1 / 2) ∫ 1 / {x √ ([1 - (x / 2) &# 178;]} DX
Let X / 2 = sinu, then x = 2sinu, DX = 2cosudu
The original formula = (1 / 2) ∫ 1 / {x √ ([1 - (x / 2) & # 178;]} DX = (1 / 2) ∫ 2cosudu / 2sinucosu = (1 / 2) ∫ Du / sinu = (1 / 2) ∫ cscudu
=(1/2)ln(cscu-cotu)+C=(1/2)ln{(2/x)-[√(4-x²)]/x}+C
Higher integral ∫ 1 / (COS &; X + Sin &; x) DX
As well as the first / (COS \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\;; 8; 8; 8; 8; 8; 8; 8; 8; 8383838383838383838383838383838383838383dx = ∫ 1 / (Tan
1-(sin²a+cos²a)(sin²a+sinacosa+cos²a)]
[1 - (Sin & # 178; a + cos & # 178; a) - 3sin & # 178; ACOS & # 178; a] how did this change from top to bottom
1-(sin²a+cos²a)(sin²a+sinacosa+cos²a)]
=1-(sin²a+sinacosa+cos²a)
=1-(1+sinacosa)
=-1/2sin2a
Sin & # 178; a + cos & # 178; a = 1 and Sin & # 178; a · cos & # 178; a = (sinacosa) &# 178;. In this way, Sin & # 178; a
Ask how to solve. Stupid will not That's the question just now
Suppose Sin & # 178; a · cos & # 178; a = (sinacosa) &# 178; = K (the K is solved by the previous sinacosa), and then add sin & # 178; a + cos & # 178; a = 1. So Sin & # 178; a, cos & # 178; a are two (two sum and two product) of the equation x & # 178; - x + k = 0. To solve the quadratic equation with one variable, the two solutions are sin & # 178; a and COS & # 178; a.ok?
Simplification of Sin & # 178; a + sinacosa / Sin & # 178; a + cos & # 178; a
Sin & # 178; a + sinacosa / Sin & # 178; a + cos & # 178; a result is Tan & # 178; a + Tana / Tan & # 178; a + 1. I'd like to ask you how to simplify it,
(Sin & # 178; a + sinacosa) / (Sin & # 178; a + cos & # 178; a) (the numerator and denominator are divided by cos & # 178; a)
=(tan²A+tanA)/(tan²A+1)
If the function f (x) = x2 + 2 (A-1) x + 2 is monotone in the interval (2,4), then the value range of real number a is______ ．
∵ the axis of symmetry of quadratic function f (x) = x2 + 2 (A-1) x + 2 is x = 1-A, the function f (x) = x2 + 2 (A-1) x + 2 is monotone function in interval (2,4), and the interval (2,4) is on the left or right side of the axis of symmetry, that is, 1-A ≥ 4, or 1-A ≤ 2, ∵ a ≥ - 1, or a ≤ - 3
Given that the function f (x) = 4x & # 178; - MX + 5 is an increasing function in the interval [- 2, + ∞), then a f (1) ≥ 25 B F (1) = 25 C f (1) ≤ 25 d f (1) > 25
This is a parabola with an opening upward. The axis of symmetry x = m / 8 ≤ - 2, m ≤ - 16
F (1) = 4-m + 5 = 9-m, because m ≤ - 16, - M ≥ 16, 9-m ≥ 25, that is, f (1) ≥ 25
Choose a
If the function f (x) = 4x & # 178; - MX + 5 is an increasing function in the interval [- 2, + ∞), then the value range of F (1) is
Obviously, the axis of symmetry of the function is x = m / 8
The function f (x) = 4x & # 178; - MX + 5 is an increasing function in the interval [- 2, + ∞)
So m / 8
It is known that the function f (x) = 4x & # 178; - MX + 15 is an increasing function in the interval [- 2, + ∞], which is the range of determining f (1)
It is known that the function f (x) = 4x & # 178; - MX + 15 is an increasing function in the interval [- 2, + ∞],
Then the axis of symmetry
x=m/8=35
RELATED INFORMATIONS
- 1. Given that the function f (x) = 4x2-mx + 5 is an increasing function in the interval [- 2, + ∞), then the value range of M is () A. （-∞，-16）B. （-∞，16]C. （-∞，-16]D. （4，16）
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